A gaseous mixture of $\mathbf{10}\mathbf{.}\mathbf{0}$ volumes ofrole="math" localid="1656949069888" ${\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{,}\mathbf{1.00}$ volume of unreacteddata-custom-editor="chemistry" ${\mathbf{O}}_{\mathbf{2}}$ , and$\mathbf{50}\mathbf{.}\mathbf{0}$ volumes of unreacted${\mathbf{N}}_{\mathbf{2}}$ leaves an engine at$\mathbf{4}\mathbf{.0}\mathbf{}\mathbf{\text{atm}}$ and 800K. Assuming that the mixture reaches equilibrium, what are

(a) the partial pressure and

(b) The concentration (in picograms per litre,$\mathbf{\text{pg}}\mathbf{/}\mathbf{\text{L}}$ ) ofdata-custom-editor="chemistry" $\mathbf{\text{CO}}$ in this exhaust gas?data-custom-editor="chemistry" $\mathbf{2}{\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{\text{CO}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}{\mathbf{K}}_{\mathbf{\text{p}}}\mathbf{=}\mathbf{1}\mathbf{.4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}$ atdata-custom-editor="chemistry" $\mathbf{800}\mathbf{.}\mathbf{\text{\hspace{0.17em}K}}$ (The actual concentration ofdata-custom-editor="chemistry" $\mathbf{CO}$ in exhaust gas is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.)

Chemical equilibrium is the state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

a)

Given reaction:

$2{\text{CO}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\text{CO}\left(\mathrm{g}\right)+{\text{O}}_2\left(\mathrm{g}\right)$

The values are:

${\mathrm{K}}_{\text{p}}=1.4\cdot {10}^{-28}\text{at}800\text{K}$

${\text{V}}_{{\text{CO}}_2}=10\text{L}$

$V_{{\text{O}}_2}=1\text{L}$

${\text{V}}_{{\text{N}}_2}=50\text{L}$

Firstly we will calculate the total volume of the gases by knowing that the amount of $\mathrm{CO}$is very small SO:

$\text{Totalvolume}={\text{V}}_{{\text{CO}}_2}+{\text{V}}_{{\text{O}}_2}+{\text{V}}_{{\text{N}}_2}$

$\text{Totalvolume}=10\text{L}+1\text{L}+50\text{L}$

$\text{Totalvolume}=61\text{L}$

Now we will calculate the equilibrium partial pressure of ${\text{CO}}_2,{\text{O}}_2$and ${\mathrm{N}}_2$ :

${\text{P}}_{{\text{CO}}_2}=\frac{\text{Volume}}{\text{Totalvolume}}\cdot \text{Totalpressure}$

${\text{P}}_{{\text{CO}}_2}=\frac{10}{61}\cdot 4\text{atm}$

${\text{P}}_{{\text{CO}}_2}=0.6557\text{atm}$

${\text{P}}_{{\text{O}}_2}=\frac{\text{Volume}}{\text{Totalvolume}}\cdot \text{Totalpressure}$

${\text{P}}_{{\text{O}}_2}=\frac{1}{61}\cdot 4\text{atm}$

${\text{P}}_{{\text{O}}_2}=0.06557\text{atm}$

${\text{P}}_{{\text{N}}_2}=\frac{\text{Volume}}{\text{Totalvolume}}\cdot \text{Totalpressure}$

${\text{P}}_{{\text{N}}_2}=\frac{50}{61}\cdot 4\text{atm}$

${\text{P}}_{{\text{N}}_2}=3.279\text{atm}$

Now we will use the reaction for the equilibrium constant to express and calculate the partial pressure of CO at equilibrium:

${\mathrm{K}}_{\text{p}}=\frac{{\left({\mathrm{P}}_{\text{CO}}\right)}^2{\mathrm{P}}_{{\text{O}}_2}}{{\left({\mathrm{P}}_{{\text{CO}}_2}\right)}^2}$

$1.4\cdot {10}^{28}=\frac{{\left(P_{\text{CO}}\right)}^2\cdot 0.06557\text{atm}}{{(0.6557\text{atm})}^2}$

${\left({\mathrm{P}}_{\text{CO}}\right)}^2=\frac{1.4\cdot {10}^{28}\cdot {(0.6557\text{atm})}^2}{0.06557\text{atm}}$

${\left({\mathrm{P}}_{\text{CO}}\right)}^2=9.1798\cdot {10}^{28}$

${\mathrm{P}}_{\text{CO}}=\sqrt{9.1798\cdot {10}^{28}}$

${\mathrm{P}}_{\text{CO}}=3\cdot {10}^{14}\text{atm}$

Therefore, the required value is ${\mathrm{P}}_{\text{CO}}=3\cdot {10}^{14}\text{atm}$.

Let us solve the given problem.

${\left[\text{CO}\right]}_{\text{equilibrium}}=?$- The concentration of $\mathrm{CO}$at equilibrium to moles per litre

It is expressed as:

data-custom-editor="chemistry" ${\left[\text{CO}\right]}_{\text{eq}}=\frac{\mathrm{n}}{\mathrm{V}}$

${\left[\text{CO}\right]}_{\text{eq}}=\frac{\mathrm{P}}{\text{RT}}$

${\left[\text{CO}\right]}_{\text{eq}}=\frac{3\cdot {10}^{14}\text{atm}}{0.0821\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\cdot 800\text{K}}$

${\left[\text{CO}\right]}_{\text{eq}}=4.61\cdot {10}^{16}\text{mol}/\text{L}\cdot \frac{28.01\text{gCO}}{1\text{molCO}}\cdot \frac{1\text{pg}}{{10}^{12}\text{g}}$

${\left[\text{CO}\right]}_{\text{eq}}=0.013\text{pg}/\text{L}$.

Therefore, the required value is${\left[\text{CO}\right]}_{\text{eq}}=0.013\text{pg}/\text{L}$ .